Electric flux through a cube problem

In summary, the cube in the figure has sides of length 10.0 cm and a uniform electric field with magnitude 4.00*10^3 N/C parallel to the xy-plane at an angle of 36.9 degrees from the +x-axis towards the +y-axis. The question asks for the electric fluxes through each of the faces and the sum. The sum is zero because the angle between the electric field and the surface is 90 degrees, resulting in a flux of zero. The electric flux through the top and bottom faces is also zero because the electric field is perpendicular to these surfaces. The electric flux through face one is not E*L^2, but rather EA*cos(theta), with theta being the angle between
  • #1
erik-the-red
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Question:

The cube in the figure (attachment) has sides of length [tex]L=10.0 {\rm cm}[/tex]. The electric field is uniform, has a magnitude [tex]E=4.00 \times 10^{3} {\rm N}/{\rm C}[/tex], and is parallel to the xy-plane at an angle of [tex]36.9^\circ[/tex]c measured from the [tex]+ x - {\rm axis}[/tex] toward the [tex]+ y - {\rm axis}[/tex].

The question asks for the electric fluxes through each of the faces and the sum.

I don't really understand why the electric flux through the top and bottom faces of the cube is zero. Is it because the angle between the face and the electric field is [tex]90^\circ[/tex]?

The sum is zero because the electric field goes in through two of the faces and then leaves through two others, right?

My answer for the electric flux through face one was [tex]E \cdot L^2[/tex] or (4.00*10^(3))*(.10^2), but that was wrong. I thought the angle was 180?
 
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  • #2
I don't really understand why the electric flux through the top and bottom faces of the cube is zero. Is it because the angle between the face and the electric field is 90?
Yes. The flux is [tex]\oint \vec{E} \cdot d\vec{A}[/tex]. The dot product can be rewritten with a cosine. Since the angle between the field and the surface is 90 degrees, the flux resolves to zero.

The electric field is piercing the surface at an angle. Draw a diagram. Since you know that [tex]\Phi = \oint \vec{E} \cdot d\vec{A}[/tex], evaluate that integral (which, in this simplified case, can be written as [tex]\Phi = EA\cos{\theta}[/tex]). The angle is not 180.
 
  • #3


I would like to clarify some concepts regarding electric flux and its calculation in this problem. Firstly, electric flux is a measure of the electric field passing through a given surface. It is defined as the dot product of the electric field and the surface area vector. In this case, the electric field is uniform and parallel to the xy-plane, meaning that the angle between the electric field and the surface area vector is always 90 degrees. Therefore, the electric flux through the top and bottom faces of the cube is indeed zero, as the angle between the electric field and the surface area vector is 90 degrees for these faces.

Secondly, the sum of the electric fluxes through all six faces of the cube is not necessarily zero. It depends on the net charge enclosed within the cube. If there is no net charge enclosed, then the sum of the electric fluxes through all six faces will be zero, as the electric field lines entering the cube will be equal to the electric field lines leaving the cube. However, if there is a net charge enclosed, the sum of the electric fluxes through all six faces will not be zero, as there will be a net electric flux passing through the cube.

Finally, the electric flux through face one is not simply the product of the electric field and the surface area. The correct formula for calculating the electric flux through a face is E*A*cos(theta), where E is the magnitude of the electric field, A is the surface area, and theta is the angle between the electric field and the surface area vector. In this case, the angle between the electric field and the surface area vector is not 180 degrees, but rather 36.9 degrees, as stated in the problem. Therefore, the correct calculation for the electric flux through face one would be (4.00*10^3)*(0.1^2)*cos(36.9 degrees).

I hope this clarifies any confusion regarding the electric flux through a cube problem. It is important to remember the definitions and formulas for electric flux and to carefully consider the angles between the electric field and the surface area vectors when calculating the electric flux through different faces.
 

1. What is electric flux through a cube?

The electric flux through a cube refers to the measure of the amount of an electric field passing through the surface of a cube. It is a useful concept in understanding the flow of electric field lines through a given surface.

2. How is electric flux through a cube calculated?

The electric flux through a cube can be calculated by taking the dot product of the electric field vector and the normal vector of the cube's surface. This is represented by the formula Φ = E∙A, where Φ is the electric flux, E is the electric field, and A is the surface area of the cube.

3. What is the unit of electric flux?

The unit of electric flux is measured in volt-meters (V⋅m) or newton-meters squared per coulomb (N⋅m^2/C).

4. How does the orientation of the cube affect the electric flux?

The orientation of the cube does not affect the electric flux as long as the surface area remains the same. This is because the electric flux is a scalar quantity and is independent of direction.

5. What are some real-world applications of understanding electric flux through a cube?

Understanding electric flux through a cube can help in various fields such as electronics, telecommunications, and energy distribution. It can also aid in the design and analysis of electrical systems, such as capacitors and conductors, and in understanding the behavior of electric fields in different materials.

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